﻿ Defining and Interpreting the Geometric Product | Primer on Geometric Algebra | David Hestenes

## Defining and Interpreting the Geometric Product

Previous section: Introduction to Geometric Algebra and Basic 2D Applications.

Next sectionRotors and Rotations in the Euclidean Plane.

The algebraic properties of vector addition and scalar multiplication are insufficient to characterize the geometric concept of a vector as a directed line segment, because they fail to encode the properties of magnitude and relative direction. That deficiency is corrected by defining suitable algebraic rules for multiplying vectors.

Geometric product:
The product ab for
vectors a, b, c is defined by the rules

• Associative:                (ab)c = a(bc)

• Left distributive:         a(c) = ab + ac

• Right distributive:      (c)a = ba + ca

• Euclidean metric:        a2 a2,

where a = |a| is a positive scalar (= real number) called the magnitude of a

In terms of the geometric product ab we can define two other products, a symmetric inner product

(1)    a= ½(ab baba

and an antisymmetric outer product

(2)    ab = ½(ab ba) = − ba

Adding (1) and (2), we obtain the fundamental formula

(3)    ab = ab + ab called the expanded form for the geometric product.

is to provide geometric interpretations for these three products.

Exercise: For a triangle defined by the vector equation a + b = c,

derive the standard Law of Cosines:  a2 b2 2ac2, and so prove that the inner product ab is always scalar-valued.

{Just multiply c2 = (a + b) (a + b) and use the Euclidean metric above and definition (1) above for the inner product.
Notice also that
if a and b are perpendicular, using the Pythagorean Theorem we can conclude ab is zero (and vice versa).
Further,
if a and b are perpendicular, using formulas (1) & (2) above, ab = ab = ½(ab − ba), which means ab = − ba.}

Therefore, the inner product can be given the usual geometric interpretation as the length of a perpendicular projection of one line segment on the direction of another.

The outer product  ab = − ba  generates a new kind of geometric quantity called a bivector, that can be interpreted geometrically as directed area in the plane of a and b.

We have shown that the geometric product interrelates three kinds of algebraic entities: scalars (0-vectors), vectors (1-vectors), and bivectors (2-vectors) that can be interpreted as geometric objects of different dimension. Geometrically, scalars represent 0-dimensional objects, because they have magnitude and orientation (sign) but no direction. Vectors represent directed line segments, which are 1-dimensional objects. Bivectors represent directed plane segments, which are 2-dimensional objects. It may be better to refer to a bivector

$\Large{{\bf{B}} = B{\bf{\hat B}}}$

as a
directed area, because its magnitude B = |B| is the ordinary area of the plane segment and its direction

$\Large{{{\bf{\hat B}}}}$

represents the plane (and orientation) in which the segment lies, just as a unit vector represents the direction of a line. The shape of the plane segment is not represented by any feature of B, as expressed in the following equivalent geometric depictions for B (with
counterclockwise orientation):

Prove the following:

Given any non-zero vector a in the plane of bivector B, one can find a vector b such that

B = ba = –ab

B2 = −|B|2 = −a2b2,

aB = –Ba,        that is, B anticommutes with every vector in the plane of B

{Looking again at the depictions directly above, it should be clear we can pick a b perpendicular to a
and also in the plane of B so that the rectangle depicting ba has the magnitude and orientation of B.

Further, since we've picked
b to be perpendicular to a, by the arguments in the exercise above,

ba = ba = – ab. Then, B2 = (ab) (ba) = – a2b2 = − a2b2 = |B|2 < 0.

Finally,
aB – aab = − a2= − baa = – Ba.}

Every vector a has a multiplicative inverse:

$\Large{{{\bf{a}}^{ - 1}} = \frac{1}{{\bf{a}}} = \frac{{\bf{a}}}{{{a^2}}}}$

that is, geometric algebra makes it possible to divide by vectors.

{Simply multiply a on either side by a/a2. As usual, we must assume a≠0.}

Prove the following theorems about the geometric meaning of commutivity and anticommutivity:

ab 0        ab = −ba          Orthogonal vectors anticommute!!          {See the exercise above.}

= λb        a0        ab ba          Collinear vectors commute!!

The problem remains to assign geometric meaning to the quantity ab without expanding it into inner and outer products.

Previous sectionIntroduction to Geometric Algebra and Basic 2D Applications.

Next section: Rotors and Rotations in the Euclidean Plane.